Digital text sets pdf

2
PREFACE
The primary objective of this work is to provide a digital text book
on SETS and from which an average student of Mathematics can
acquire as much depth and comprehension in their study of
mathematics. This text is an outline for sets and can be used as a
reference.
The text mainly consists of three chapters. Chapter 1 deals about
sets and their representations. Chapter 2 deals with the graphical
representation of sets ,Venn diagrams ,and operations on sets .
Chapter 3 is giving a number of practical life problems related to
sets.
Each chapter consists of definitions, examples and number of
exercises are provided.

3
CONTENTS
CHAPTER 1 SETS 4 -11
CHAPTER 2 VENN DIAGRAM 12 – 18
CHAPTER 3 PRACTICAL PROBLEMS
ON UNION AND INTERSECTION 18 - 23
REFERENCE 24
ANSWERS 25

4
CHAPTER 1
SETS
1.1 Introduction
The concept of set serves as a fundamental part of the present day
mathematics. Today this concept is being used in almost every branch of
mathematics .Sets are used in define the concepts of relations and
functions. The study of geometry , sequences , probability , etc requires
the knowledge of sets.
The theory of sets was developed by German mathematician Georg cantor
. He first encountered sets while working on “ problems on trigonometric
series”.In this chapter we discuss some basic definitions and types of sets.
1.2 Sets and their representations
A set is a well defined collection of objects
The following points may be noted.
(i) Objects , elements and members of a set are synonymous terms.
(ii) Sets are usually denoted by capital letters A , B ,C , X, Y, Z
(iii) The elements of a set are represented by small letters a ,b, c, x, y, z
We give below a few more examples of sets used particularly in
mathematics:
ℕ: 𝑡𝑕𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
ℤ: 𝑡𝑕𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑖𝑛𝑡𝑒𝑔𝑒𝑟𝑠
ℝ ∶ 𝑡𝑕𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
ℂ ∶ 𝑡𝑕𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑐𝑜𝑚𝑝𝑙𝑒𝑥 𝑛𝑢𝑚𝑏𝑒𝑟
ℚ: 𝑡𝑕𝑒 𝑠𝑒𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑟𝑎𝑡𝑖𝑜𝑛𝑎𝑙𝑠
ℝ+
: the set of all positive rationals
If a is an element of a set A ,we say that “ a belongs to A” the Greek
symbol ∈ is used to denote the phrase „belongs to‟. Thus we write a ∈A .If
„b‟is not an element of a set A,we write 𝑏 ∉ 𝐴 and read “b does not belongs
to A”.

5
Thus , in the set V of vowels in the English alphabet , a∈V but b ∉ 𝑉
There are two methods of representing a set:
(i) Roster or tabular form
(ii) Set – builder form
In Roster form , all the elements of a set are listed , the
elements are being separated by commas and are enclosed within braces
{ } . For example , the set of all even positive integers less than 7 is
described in rosters form as {2,4,6}.Some more examples of representing
a set in roster form are given below:
(a) The set of all natural numbers which divide 42 is { 1,2,3,6,7,14,21,42}
(b)The set of all vowels in English alphabet is {a,e,i.o.u}
(c) The set of all odd numbers is represented by {1,3,5…}.The dots
denotes that the list of odd numbers continue indefinitely
(d) In set builder form , all the elements of a set possess a single
common property which is not possessed by any element outside the
set. For example , in the set {a,e,i.o.u},all the elements possess a
common property, namely each of them is a vowel in the English
alphabet , and no other letter possess this property.Denoting this set by
V , we write
V = { x: x is a vowel in English alphabet}
Consider the set A = { x: x is a natural number and 3 < x < 10} is read
as “ the set of all x such that x is a natural number and x lies between 3
and 10”.Hence , the numbers 4, 5,6,7,8 and 9 are the elements of the set
A.
Example 1 Write the solution set of the equation x2
+x-2=0 in roster
form
Solution The given equation can be written as
(x-1) ( x +2)=0
x=1 , x=2
Therefore, the solution set of the given equation can be written in roster
form as {1, 2}

6
Example 2 Write the set {x : x is a positive integer and x2
< 40} in
roster form
Solution {1,2,3,4,5,6}
Example 3 Write the set A= {1,4,9,16,25,…} in set builder form
Solution We may write the set A as
A= { x: x is the square of a natural number}
EXERCISE 1.1
1. Write the following sets in roster form
(i) A = {x : x is an integer and -3 < x< 7}
(ii) B = { x : x is a natural number less than 6 }
(iii) C = The set of all letters in the word TRIGNOMETRY
(iv) D ={ x : x is a prime number which is divisor of 60}
2. Write the following sets in the set builder form
(i) {3,6,9,12}
(ii) {2,4,8,16,32}
(iii) { 2,4,6 …}
(iv) {5,25,125,625}
3. Let A = {1,2,3,4,5,6}.Insert the appropriate symbol ∈ and ∉ in the
blank spaces
(i) 5 A (ii) 8 A (iii) 3 A (iv) 6 A
1.3The empty set
Consider the set B ={ x: x is student studying in both classes X and XI}.
We observe that a student cannot study simultaneously in both classes X
and XI . Thus , the set B contains no element al all.
Definition 1 A set which does not contain an element is called empty
set or the null set.
According to this B is an empty set and is denoted by ∅ or { }.
We give below a few examples of empty sets:
(i) Let A = {x : 1 < x < 2 ,x is a natural number }. Then A is
the empty set , because there is no natural number between
1 and 2
(ii) B= { x : x is an even prime number greater than 2}. Then
Bis the empty set because 2 is the only even prime number.

7
1.4 Finite and Infinite Sets
Let A ={1,2,3,4,5} B={ a, b, c, d, e, g} and C= { men living presently in
different parts of the world}.
We observe that A contains 5 elements and B contains 6 elements . How
many elements does C contain? As it is , we do not know the number of
elements in C, but it is some natural number which may be quite a big number
.By number of elements in a set S , we mean the number of distinct elements
of the set and denote it by n(S).If n(S) is a natural number , then S is non
empty finite set.
Consider the set of natural numbers , we see that the number of elements
of this set is not finite since there are infinite number of natural numbers and
we say that the set of natural numbers is an infinite set. The sets A, B and C
given above are finite sets and n (A)=5 ,n(B) =6 and n(C)=some finite
number.
Definition 2 A set which is empty or consists of a definite number of
elements is called finite otherwise ,the set is called infinite.
Example 4 State which of the following sets are finite or infinite
(i) {x: x ∈ ℕ 𝑎𝑛𝑑 𝑥 − 2 𝑥 − 1 = 0}
(ii) { 𝑥: 𝑥 ∈ ℕ 𝑎𝑛𝑑 𝑥2
= 4}
(iii) {𝑥 ∶ 𝑥 ∈ ℕ 𝑎𝑛𝑑 2𝑥 − 1 = 0 }
(iv) {𝑥 ∶ 𝑥 ∈ ℕ 𝑎𝑛𝑑 𝑥 𝑖𝑠 𝑝𝑟𝑖𝑚𝑒}
(v) {𝑥 ∶ 𝑥 ∈ ℕ 𝑎𝑛𝑑 𝑥 𝑖𝑠 𝑜𝑑𝑑}
Solution
(i) Given set ={1,2}.Hence it is finite.
(ii) Given set = {2}. Hence it is finite.
(iii) Given set = Φ.Hence it is finite.
(iv) The given set is the set of all prime numbers and since the
set of prime numbers is infinite . Hence the given set is
infinite.

8
(v) Since there are infinite number of odd numbers , hence the
given set is infinite.
1.5 Equal Sets
Definition 3 Two sets A and B are said to be equal if they have exactly
the same elements and we write A=B .Otherwise , the sets are said to be
unequal and we write A≠B.
We consider the following examples;
(i) Let A ={1,2,3,4} and B ={3,1,4,2}.Then A=B.
(ii) Let A be the set of prime numbers less than 6 and P the set of prime
factors of 30. Then A and P are equal , since 2,3 and 5 are the only
prime factors of 30 and these are less than 6.
Example 5 Find the pair of equal sets , if any give reasons
A = {0} B={𝑥: 𝑥 > 15 𝑎𝑛𝑑 𝑥 < 5}
C = { 𝑥: 𝑥 − 5 = 0} D = {𝑥: 𝑥2
= 25}
E={𝑥: 𝑥 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑟𝑜𝑜𝑡 𝑜𝑓 𝑡𝑕𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑥2
− 2𝑥 − 15 = 0}
Solution Since 0 ∈ A and 0 does not belongs to any set of the sets B, C, D
and E ,it follows that A≠ B , A ≠C , A ≠ E, A ≠ E
Since B = ∅ but none of the other sets are empty. Therefore B≠ C ,B≠ D ,
B ≠E .Also C ={5} but -5 ∈ D, hence C ≠D.
Since E = {5}, C=E .Further , D= {-5,5} and E={5}.We find that D≠ E. Thus
,the only pair of equal set is C and E
1.6Subsets
Consider the sets : X= set of all students in your school, Y= set of all
students in your class.
We note that every element of Y is also an element of X. The fact that Y
is a subset of X is expressed in symbols as Y⊂ X. The symbol ⊂ stands
for „is a subset of ‟or „is contained in‟
Definition 4 A set A is said to be a subset of a set B if every element of A
is also an element of B.

9
In other words , A⊂B if whenever a ∈ 𝐴 ,then a ∈B.It is convenient to
use the symbol " ⇒ "which means implies .Using this symbol , we can
write the definition of subset as follows
A ⊂ B 𝑖𝑓 𝑎 ∈ 𝐴 ⇒ 𝑎 ∈ 𝐵
We read the above statement as “A is a subset of B if a is an element of A
implies that a is also an element of B”
Now , A ⊂ B means every element of A is in B and every element of B
may or may not be in A . If so happens that every element of B is in A
,then A and B are the same sets and vice versa.
That is ,A ⊂ B and B ⊂A ⇔ A=B(where the symbol“⇔ "𝑓𝑜𝑟 two-
way implications and is read as „if and only if‟ ,shortly „iff‟ )
It follows from the above definition that every set A is a subset of itself
i.e., A⊂ A .Since the empty set has no elements , we can say that ∅ is a
subset of every set.
We consider the following examples:
(i) The set ℚ of rational numbers is a subset of the set ℝ of real
numbers.
(ii) Let A= {1,3,5}and B={x: x is an odd natural number less than
6}.Then A⊂ B and B⊂ A ,Hence A =B
(iii) Let A ={a,e,i,o,u} and B={a,b,c,d}.Then A is not a subset of B , also
B is not a subset of A.
Let A and B be two sets .If A⊂B and A ≠B ,then A is called a proper
subset of B and B is called superset of A.F or example,
A ={2,4,6} is a proper subset of B={1,2,3,4,5,6}.
If a set has only one element, we call it a singleton set. Thus {a} is a
singleton set.
Example 6 Consider the sets
∅ , A ={1,3} , B= {1,5,9} C ={1,3,5,7,9}.

10
Insert the symbol ⊂ or ⊄ between each of the following pair of sets
(i) ∅ … B (ii) A…B (iii) A … C (iv) B… C
Solution
(i) ∅ ⊂ B as ∅ is subset of every set .
(ii) A⊄ B as 3∈ A and 3∉ B
(iii ) A ⊂ C as 1,3, ∈ A also belongs to U
(iv) B ⊂ C each element of B is also an element of C.
Subsets of sets of real numbers
There are many important subsets of ℝ.Some of them are given below;
The set of natural numbers ℕ = {1,2,3,4,5…}
The set of integers ℤ = {…,-3,-2,-1,0,1,2,3…}
The set of rational numbers ℚ = {𝑥: 𝑥 =
𝑝
𝑞
p,q,∈ ℤ}
The set of irrational numbers , denoted by 𝚻 composed of all other real
numbers. Thus 𝚻 = {𝑥: 𝑥 ∈ ℝ 𝑎𝑛𝑑 𝑥 ∉ ℚ }.That is all real numbers that are
not rational.
Some of the obvious relations among these subsets are
ℕ ⊂ ℤ ⊂ ℚ , ℚ ⊂ ℝ , 𝚻 ⊂ ℝ
1.7Power Set
Consider the set {1,2}. Now let us write all the subsets of {1,2}.It has
four subsets in all viz. ∅ , {1,2},{1},{2}.The set of all these subsets is
called the power set of {1,2}.
Definition 5 The collection of all subsets of a set A is called the power
set of A .It is denoted by P(A).In P(A) , every element is a set.
Thus as in above , if A ={1,2},then
P(A) = {∅, 1,2 , 1 , {2}}

11
Also note that n[P(A)] =4=22
In general , if A is a set with n(A)=m ,then n[P(A)]= 2 𝑚
1.8 Universal set
Usually , in a particular context ,we have to deal with the elements and
subsets of a basic set which is relevant to that particular context. For
example, while studying the system of numbers ,we are interested in the set
of natural numbers and its subsets such as the set of all prime numbers , the
set of all even numbers ,and so forth. This basic set is called the “Universal
Set” . This universal set is usually denoted by U ,and all its subsets by the
letters A,B,C etc.
For example , for the set of all integers, the universal set can be the set of
all rational numbers, or ,for that matter ,the set of all real numbers. For
another example , in human population studies, the universal set consist of all
the people in the world.
EXERCISE 1.2
1 . Which of the following sets are finite or infinite
(i) The set of months of an year
(ii) {1,2,3,…}
(iii) {1,2,3,…99,100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
2 . Are the following pairs of sets equal ? Give reasons.
(i) A={2,3} B={𝑥: 𝑥 𝑖𝑠 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥2
+ 5𝑥 + 6 = 0}
(ii) A ={𝑥: 𝑥 𝑖𝑠 𝑎 𝑙𝑒𝑡𝑡𝑒𝑟 𝑖𝑛 𝑡𝑕𝑒 𝑤𝑜𝑟𝑑 𝐹𝑂𝐿𝐿𝑂𝑊}
B={𝑦: 𝑦 𝑖𝑠 𝑎 𝑙𝑒𝑡𝑡𝑒𝑟 𝑖𝑛 𝑡𝑕𝑒 𝑤𝑜𝑟𝑑 𝑊𝑂𝐿𝐹}
3 . Write down all the subsets of the following sets
(i) {a} (ii) {a , b} (iii) {1,2,3,4,5} (iv) ∅
4 . How many elements has P(A) , if A= ∅ ?

12
CHAPTER 2
VENN DIAGRAMS
2.1 Introduction
Most of the relationships between sets can be represented by means of
diagrams which are known as Venn diagrams. Venn diagrams are named
after the English logician , John Venn (1834-1883).These diagrams consist
of rectangles and closed curves usually circles. The universal set is
represented by a rectangle and its subsets by circles.
In Venn diagrams, the elements of the sets are written in their
respective circles( fig.2.1)
(fig 2.1)
Illustration
In the fig 2.1 , U={1,2,3…,10} is the universal set of which
A ={2,4,6,8,10} is a subset.
2.2 Operations on Sets
Here we define certain operations on sets and examine their
properties.
2.2.1 Union of Sets Let A and B be any two sets. The union of A and B is
the set which consists of all the elements of A and all the elements of B, the
common elements being taken only once. The symbol “∪ " is used to denote
the union. Symbolically we write A ∪ B and usually read as “A union B”
• 1 A • 3
• 5
• 9 • 7
•2
• 4 • 8
• 6 •10

13
Example 7 Let A={2,4,6,8} and B ={6,8,10,12}. Find A ∪ B
Solution We have A ∪ B={2,4,6,8,10,12}
Example 8 Let A= {a ,e ,i ,o ,u} and B ={a, i ,u} .Show that A ∪ B=A.
Solution We have A ∪ B={a ,e ,i, o,u}=A.
This example illustrate that the union of sets A and its subset B is the set A
itself. That is , if B ⊂ A ,then A ∪ B= A.
Definition 6 The union of two sets A and B is the set U which consists of all
those elements which are either in A or in B.
In symbols , we write A ∪ B={𝑥: 𝑥 ∈ 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}.
Fig 2.2
The shaded portion in Fig 2.2 represents A ∪ B.
Some properties of the Operation of Union
(i) A ∪ B= B ∪ A.( Commutative law)
(ii) (A ∪ B.) ∪ 𝐶 = A∪ (𝐵 ∪ C) (Associative law)
(iii) 𝐴 ∪ ∅ = 𝐴 (Law of identity element)
(iv) 𝐴 ∪ 𝐴 = 𝐴 (Idempotent law)
(v) 𝑈 ∪ 𝐴 = 𝑈 (Law of U)

14
2.2.2 Intersection of Sets The intersection of sets A and B is the set of all
elements which are common to both A and B. The symbol “∩”is used to
denote the intersection. The intersection of two sets A and B is the set of all
those elements which belong to both A and B. Symbolically , we write
A ∩ B = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵}
Example 9 Consider the sets A and B of example 7
Solution We see that 6,8 are the only elements which are common to both A
and B. Hence A ∩ B = {6, 8}.
Example 10 Let A ={1,2,3,4,5,6,7,8,9,10} and B ={2,3,5,7}. Find A ∩ B and
hence show that A ∩ B=B
Solution We have A ∩ B={2,3,5,7}=B.
Definition 7 The intersection of two sets A and B is the set of all those
elements which belong to both A and B. Symbolically we write
A ∩ B ={𝑥 ∶ 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵}.
The shaded portion in Fig 2.3 indicates the intersection of A and B.
Fig 2.3
If A and B are two sets such that A ∩ B =∅ , then A and B are called disjoint
sets.

15
For example , let A={2,4,6,8} and B ={1,3,5,7}.Then A and Bare disjoint
sets, because there are no elements which are common to A and B.The
disjoint sets can be represented by means of Venn diagram as shown in Fig
2.4
Fig 2.4
In the above diagram A and B are disjoint sets.
Some properties of Operation of Intersection
(i) A ∩ B = B ∩ A (Commutative law)
(ii) (A ∩ 𝐵) ∩ C = A ∩ (B ∩ C) (Associative law)
(iii) ∅ ∩ A = ∅ ,U ∩ A = A (Law of ∅ and U)
(iv) 𝐴 ∩ A= A (Idempotent law)
(v) 𝐴 ∩ (B ∪ C) =(A ∩ 𝐵) ∪ (A ∩ C ) (Distributive law).
2.2.3 Difference of Sets
The difference of the sets A and B in this order is the set of elements
which belong to A but not to B. Symbolically , we write 𝐴 − 𝐵 and read as
“A minus B”.
Example 11 Let A ={1,2,3,4,5,6} , B = {2,4,6,8}.Find 𝐴 − 𝐵 𝑎𝑛𝑑 𝐵 − 𝐴.
Solution We have , 𝐴 − 𝐵 ={1,3,5} , since the elements 1,3,5 belong to A
but not to B and 𝐵 − 𝐴 = {8} ,since the element 8 belong to B but not to A.
We note that 𝐴 − 𝐵 ≠ 𝐵 − 𝐴
Example 12 Let V={a,e,i,o,u} and B={a,i,k,u} .Find 𝑉 − 𝐵 𝑎𝑛𝑑 𝐵 − 𝑉.
Solution We have 𝑉 − 𝐵 ={e,o} ,since the elements e,o belong to V but not
to B and 𝐵 − 𝑉={k} ,since the element k belong to B but not to V.
U
A B

16
We note that 𝑉 − 𝐵 ≠ 𝐵 − 𝑉.Using the set builder notation ,we
can rewrite the definition of difference as
𝐴 − 𝐵 = {𝑥: 𝑥 ∈ 𝐴 𝑎𝑛𝑑 𝑥 ∉ 𝐵}.
The difference of two sets A and B can be represented
by Venn diagram as shown in Fig.2.5
Remark The sets 𝐴 − 𝐵, 𝐴 ∩ 𝐵, 𝐵 − 𝐴 are mutually
disjoint sets, i.e., the intersection of any of these two
sets is the null set.
2.3 Complement of a set
Let U be the universal set which consists of all prime numbers and A be
the subset of U which consists of all those prime numbers that are not divisors
of 42.Thus A={𝑥: 𝑥 ∈ 𝑈 𝑎𝑛𝑑 𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑑𝑖𝑣𝑖𝑠𝑜𝑟 𝑜𝑓 42}.We see that 2∈ 𝑈
but 2∉ A, because 2 is a divisor of 42.Similarly 3 ∈ 𝑈 , 𝑏𝑢𝑡 3 ∉ A ,and 7 ∈ 𝑈
but 7∉ A. Now 2,3, and 7 are the only elements of U which do not belong to
A. The set of these three prime numbers , i.e., the set {2,3,7} is called the
complement of A with respect to U ,and is denoted by A'. So we have
A'={2,3,7}.Thus we see that
A'={𝑥: 𝑥 ∈ 𝑈 𝑎𝑛𝑑 𝑥 ∉ 𝐴}.
Definition 8 Let U be the universal set and A a subset of U. Then the
complement of A is the set of all elements of U which are not the elements of
A. Symbolically we write A' to denote the complement of A with respect to
U. Thus ,
A'={𝑥: 𝑥 ∈ 𝑈 𝑎𝑛𝑑 𝑥 ∉ 𝐴}.Obviously A'=𝑈 − 𝐴
Example 13 Le U={1,2,3,4,5,6,7,8,9,10} and A={1,3,5,7,9}.Find A'
Solution A'={2,4,6,8,10}, Since 2,4,6,8,10 are the only elements of U which
do not belong to A.

17
Note: If A is a subset of the universal set U, then its complement A' is also a
subset of U.
Again in example 13 above, we have A'={2,4,6,8,10}.
Hence (A')'={𝑥: 𝑥 ∈ 𝑈 𝑎𝑛𝑑 𝑥 ∉ A′}
={1,3,5,7,9}
=A
That is (A')'=A
Example 14 Let U={1,2,3,4,5,6} , A={2,3} and B={3,4,5}.Find A' , B',
A'∩ B' , A ∪ B , and hence show that (A ∪ B)' = A'∩ B'.
Solution Clearly A'={1,4,5,6} , B'={1,2,6}. Hence A'∩ B' ={1,6}
Also A ∪ B={2,3,4,5}
(A ∪ B)'={1,6} = A'∩ B'.
It can be shown that the above result is true in general. If A and B are any two
subsets of the universal set U, then
(A ∪ B)' = A'∩ B'.
Similarly , (A ∩B )' = A'∪ B'. These two results are stated in words as
follows:
The complement of the union of two sets is the intersection of their
complements and the complement of the intersection of two sets is the union
of their complements. These are called Demorgan‟s laws. These are named
after the mathematician De Morgan.
The complement A' of a set A can be represented by a Venn diagram as
shown in Fig 2.6

18
Some properties of complement Sets
(i) Complement laws : A ∪ A'=U , A ∩ A' = ∅
(ii) Demorgan‟s law : (A ∪ B)' = A'∩ B' , (A ∩B )' = A'∪ B'
(iii) Law of double complementation (A')'=A
(iv) Laws of empty set and universal set ∅′
=U and U '=∅
EXERCISE 2.1
(i) Let A = {a ,b} B ={a,b,c} .Is A ⊂ B ? What A ∪ B ?
(ii) If A= {1,2,3,4} B={3,4,5,6} C= {5,6,7,8} and D ={7,8,9,10}.Find
(i) A ∪ B (ii) A ∪ C (iii) B ∪ C (iv) B ∪ D
(iii ) If ℝ is the set of real numbers and ℚ is the set of rational numbers ,
then what is ℝ − ℚ ?
(iv ) If X={a,b,c,d} and Y={f,b,d,g} .Find
(i) X – Y (ii) Y – X (iii) X ∩ Y
(v) Draw appropriate Venn diagrams for the following
(i) (A ∪ B )'
(ii) A' ∩ B'
(iii) (A∩ B)'
(iv) A' ∪ B'

19
CHAPTER 3
PRACTICAL PROBLEMS ON UNION AND INTERSECTION
Introduction
In second chapter, we have learnt union, intersection, and difference of
two sets. In this chapter , we will go through
some practical problems related to our daily
life.
Note the following formulae derived.
Let A and B be finite sets. If 𝑨 ∩ 𝑩 = ∅, then
(i) n (A ∪ B) = n ( A ) + n ( B ) (1)
The elements in A ∪ B are either in A or in B but not in both as 𝑨 ∩ 𝑩 = ∅.
So (1) follows immediately.
(ii) n (A ∪ B) = n ( A ) + n ( B ) –n (A∩ B)
(2)
Note that the sets 𝐴 − 𝐵, 𝐴 ∩ 𝐵, 𝑎𝑛𝑑 𝐵 − 𝐴 are disjoint and their
union is A ∪ B.Therefore
n (A ∪ B) = n (𝐴 − 𝐵) + n (𝐴 ∩ 𝐵) +n (𝐵 − 𝐴 )
= n (𝐴 − 𝐵) + n (𝐴 ∩ 𝐵) +n (𝐵 − 𝐴 ) + n(A∩ B) − n(A∩ B)
= n (A) +n (B) − n(A∩ B) , which verifies (2)
(iii) If A ,B, and C are finite sets , then
n (A∪ B ∪ C) = n (A) + n (B) + n (C) − n(A∩ B) – n (B∩ C)
– n (A∩ C) + n (A ∩ B ∩ C) (3)
Example 15 If X and Y are two sets such that X ∪ Y has 50 elements , X has
28 elements and Y has 32 elements , how many elements does X ∩ Y have ?

20
Solution Given that n(X ∪ Y) = 50 n(X) =28 n(Y) = 32
By using the formula ,
n (X ∪ Y) = n(X) + n(Y) – n (X ∩ Y)
we find that
n (X ∩ Y) = n(X) + n(Y) – n (X ∪ Y)
= 28+32 – 50
= 10
Example 16 In a school there are 20 teachers who teach mathematics or
physics. Of these , 12 teach mathematics and 4 teach both physics and
mathematics. How many teach Physics ?
Solution Let M denote the set of teachers who teach mathematics and P
denote the set of teachers who teach Physics. In the statement of the problem
, the word “ or ” gives us a clue of union and the word “and ” gives us a clue
of intersection. We therefore ,have
n (M ∪ 𝑃 ) =20, n (M) =12 and n( M ∩P) =4.
We wish to determine n (P)
n (M ∪ 𝑃 ) = n (M) + n (P) – n( M ∩P)
20 = 12 + n (P) – 4
n (P) = 20-12+4
= 12
Hence 12 teachers teach Physics.
Example 17 In a class of 35 students , 24 like to play cricket and 16 like to
play football . Also , each student likes to play at least one of the two games.
How many students like to play both cricket and football?

21
Solution Let X be the set of students who like to play cricket and Y be the
set of students who like to play football. Then X ∪ Y is the set of students
who like to play at least one game , and X ∩ Y is the set of students who like
to play both games.
Given n(X) =24 n(Y) =16 n (X ∪ Y) =35 n (X ∩ Y) =?
Using the formula
n (X ∪ Y) = n(X) + n(Y) – n (X ∩ Y)
35 = 24 + 16 – n (X ∩ Y)
n (X ∩ Y) = 24+ 16 – 35
= 5
That is 5 students like to play both games.
Example 18 In a survey of 400 students in a school , 100 were listed as taking
apple juice , 150 as taking orange juice and 75 were listed as taking both
apple as well as orange juice . Find how many students were taking neither
apple juice nor orange juice ?
Solution Let U denote the set of surveyed students and A denote the set of
students taking apple juice and B denote the set of students taking orange
juice. Then
n(U) =400 n(A)=100 n(B)= 150 and n(A ∩ B) = 75
Now , n(A′ ∩ 𝐵′) = n( A∪B)'
= n(U) – n( A∪B)
= n(U) – n(A) – n(B) + n(A ∩ B)
= 400 – 100 – 150 +75
= 225
Hence 225 students were taking neither apple juice nor orange juice.

22
Example 19 There are 200 individuals with a skin disorder , 120 had been
exposed to the chemical C1 , 50 to chemical C2 , and 30 to both C1 and
C2.Find the number of individuals exposed to
(i) Chemical C1 but not to chemical C2
(ii) Chemical C2 but not to chemical C1
(iii) Chemical C1 or chemical C2
Solution Let U denote the universal set consisting of individuals suffering
from the skin disorder , A denote the set of individuals exposed to the
Chemical C1 and B denote the set of individuals exposed to chemical C2
Hence n(U) =200 n(A) =120 n(B) = 50 and n (A∩ B) =30
(i) We have A = 𝐴 − 𝐵 ∪ (𝐴 ∩ 𝐵) ( refer Venn diagram in chapter 2)
n (A) = n 𝐴 − 𝐵 + 𝑛 (𝐴 ∩ 𝐵)
n 𝐴 − 𝐵 = n (A) − 𝑛 (𝐴 ∩ 𝐵)
= 120 -30
= 90
(ii) Also B = (𝐵 − 𝐴) ∪ (𝐴 ∩ 𝐵)
n ( B ) = n (𝐵 − 𝐴) + 𝑛(𝐴 ∩ 𝐵)
n (𝐵 − 𝐴) = n ( B ) − 𝑛(𝐴 ∩ 𝐵)
= 50 − 30
=20
Thu the number of individuals exposed to chemical C2 and not to chemical
C1 is 20.
(iii) The number of individuals exposed either to chemical C1 or to
chemical C2 , i.e.,
n (𝐴 ∪ 𝐵) = n (A) + n (B) – n (A ∩ 𝐵 )
= 120 + 50 − 30
= 140

23
Example 20 If X and Y are two sets such that n( X)= 17 , n ( Y) = 23 and
n (X ∪ Y) = 38 , find n (X ∩ Y)
Solution n (X ∪ Y) = n( X) + n ( Y) − n (X ∩ Y)
n (X ∩ Y) = n( X) + n ( Y) − n (X ∪ Y)
= 17 + 23 – 38
= 2
EXERCISE 3.1
(i) If X and Y are two sets such that X ∪ Y 18 elements , X has 8
elements and Y has 15 elements ; how many elements does
X ∩ Y have ?
(ii) In a group of 400 people , 250 can speak Hindi and 200 can speak
English. How many people can speak both Hindi and English ?
(iii) If S and T are two sets such that S has 21 elements , T has 32
elements , and S ∩ T has 11 elements , how many elements does
S ∪ T have ?
(iv) In a group of 70 people , 37 like coffee , 52 like tea and each person
likes at least one of the two drinks . How many people like both
coffee and tea ?

24
REFERENCE
(i) Plus one text book, State Council of Research and Training
(SCERT)

25
ANSWERS
EXERCISE 1.1
1. (i) A ={-3,-2,-1,0,1,2,3,4,5,6} (ii) B={1,2,3,4,5}
(iii) C = {T ,R,I,G,N,O,M,E ,Y} (iv) D={2,3,5}
2. (i) {𝑥 ∶ 𝑥 = 3 𝑛
, 𝑛 ∈ 𝑁 𝑎𝑛𝑑 1 ≤ 𝑛 ≤ 4} (ii) {𝑥 ∶ 𝑥 = 2 𝑛
, 𝑛 ∈
𝑁 𝑎𝑛𝑑 1 ≤ 𝑛 ≤ 5} (iii) {𝑥 ∶ 𝑥 𝑖𝑠 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑛𝑢𝑚𝑏𝑒𝑟} (iv) {𝑥 ∶ 𝑥 =
5 𝑛
, 𝑛 ∈ 𝑁 𝑎𝑛𝑑 1 ≤ 𝑛 ≤ 4}
3. (i) ∈ (ii) ∉ (iii) ∈ (iv) ∉
Exercise 1.2
1. (i) Finite (ii) Infininite (iii) Finite (iv) Infinite (v) Finite
3. (i) {a} ,∅ (ii) {a} ,{b}, {a,b}, ∅ (iii) {1},{2}{3 {4} {5} {
1,2},{1,3},{1,4,},{1,5},{2,3},{2,4},{2,5},{3,1},{3,2},{3,4},{3,5},{4,1},{4,2
},{4,3},{4,5}{5,1},{5,2}{5,3},{5,4},{1,2,3,4,5},∅
4. 1
Exercise 2.1
1. (i) Yes , {a,b,c}
2. (i) {1,2,3,4,5,6} (ii){ 1,2,3,4,5,6,7,8} (iii) {3,4,5,6,7,8} (iv)
{3,4,5,6,7,8,9,10}
3. Irrational numbers
4. (i) {a,c} (ii) {f,g} (iii) {b,d}
EXEXCISE 3.1
(i) 5 (ii) 50 (iii) 42 (iv) 19

Digital text sets pdf

More Related Content

What's hot (16)

Similar to Digital text sets pdf (20)

Recently uploaded (20)

Digital text sets pdf